I have $X_1,X_2 ... X_n$ Gamma-distributed r.v with density:

enter image description here

First of all I showed, that Gamma distribution belongs to exponential family and can be represented in form

enter image description here

I found, that for Gamma distribution $B(\theta) = log(\theta)$ where $\theta$ is $\alpha/\beta$

Then we know, that moment generating function of the random variable $T(X)$ is given by $$M(t)=\exp\{B(\theta+t)-B(\theta)\}$$

Natural sufficient statistics for Gamma distribution is:

enter image description here

But I didn't understand what should I do next? Because Natural sufficient statistics is two dimensional it confuses me:

$M(T) = exp[log(\theta + T) - log(\theta)]$

and $T = [s,t]^\intercal$

  • 1
    $\begingroup$ Your representation of an exponential distribution is (a) unidimensional and (b) based on $y$ being the sufficient statistic. Check the literature or even the Wikipedia page for proper definitions. $\endgroup$ – Xi'an Dec 6 '17 at 18:15

As already recalled in the answer to a previous question of yours and well-explained on Wikipedia, for any exponential family, there exists a parameterisation such that the density of the family is$$f(x|\theta)=\exp\{\theta\cdot T(x)-\Psi(\theta)\}$$wrt a constant measure $\text{d}\mu(x)$, where the components of $T(\cdot)$ are linearly independent. The $\cdot$ in $\theta\cdot T(x)$ represents the scalar product, which means both $\theta$ and $T(x)$ are vector of the same dimension.

When you consider the density of the $\text{Ga}(\alpha,\beta)$ distribution,$$f(x|\alpha,\beta)=\dfrac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}\exp\{-\beta x\}\mathbb{I}_{(0,\infty)}(x)$$moving terms into the exponential leads to \begin{align*} f(x|\alpha,\beta)&=\dfrac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}\exp\{-\beta x\}\mathbb{I}_{(0,\infty)}(x)\\ &=\exp\{\log[\beta^\alpha] - \log \Gamma(\alpha)+[\alpha-1]\log x-\beta x\}\mathbb{I}_{(0,\infty)}(x)\\ &=\exp\{\underbrace{[\alpha-1]\log x-\beta x}_{\theta\cdot T(x)} + \underbrace{\log[\beta^\alpha] - \log \Gamma(\alpha)}_{\Psi(\theta)}\}\mathbb{I}_{(0,\infty)}(x)\\ \end{align*} which suggests $\theta=(\alpha-1,\beta)$ as natural parameter and $T(x)=(\log x,-x)$ as sufficient statistic. (But other choices are possible.)

Hence, for this parameterisation of the $\text{Ga}(\alpha,\beta)$ distribution as an exponential family, \begin{align*} \Psi(\theta)&=-\log[\beta^\alpha] + \log \Gamma(\alpha)\\ &=-(\alpha-1+1)\log \beta + \log \Gamma(\alpha-1+1)\\ &= -[1+\theta_1]\log \theta_2 +\log \Gamma(1+\theta_1) \end{align*} as the cumulant moment function.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.